Question

tính

a, A = \(\dfrac{3}{2}\) + \(\dfrac{3}{10}\) + \(\dfrac{3}{50}\)+ \(\dfrac{3}{250}\) + \(\dfrac{3}{1250}\)

b, B = \(\dfrac{3}{2.5}\)+ \(\dfrac{3}{5.8}\) + \(\dfrac{3}{8.11}\) + \(\dfrac{3}{11.14}\)

c, C = \(\dfrac{5}{2}\) + \(\dfrac{5}{6}\) + \(\dfrac{5}{12}\) + \(\dfrac{5}{20}\) + \(\dfrac{5}{30}\) + \(\dfrac{5}{42}\) + ........... + \(\dfrac{5}{110}\)

d, D = \(\dfrac{1}{2.3.4}\) + \(\dfrac{1}{3.4.5}\) + \(\dfrac{1}{4.5.6}\) + \(\dfrac{1}{5.6.7}\) + \(\dfrac{1}{6.7.8}\) + \(\dfrac{1}{7.8.9}\) + \(\dfrac{1}{8.9.10}\)

Answer 1

Đăng ít thôi.

d) \(D=\dfrac{1}{1.2.3}+\dfrac{1}{3.4.5}+\dfrac{1}{4.5.6}+\dfrac{1}{5.6.7}+\dfrac{1}{6.7.8}+\dfrac{1}{7.8.9}+\dfrac{1}{8.9.10}\)

\(\Rightarrow2D=\dfrac{2}{1.2.3}+\dfrac{2}{3.4.5}+\dfrac{2}{4.5.6}+\dfrac{2}{5.6.7}+\dfrac{2}{6.7.8}+\dfrac{2}{7.8.9}+\dfrac{2}{8.9.10}\)

\(\Rightarrow2D=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+\dfrac{1}{4.5}-\dfrac{1}{5.6}+...+\dfrac{1}{8.9}-\dfrac{1}{9.10}\)

\(\Rightarrow2D=\dfrac{1}{2.3}-\dfrac{1}{9.10}\)

\(\Rightarrow2D=\dfrac{22}{45}\)

\(\Rightarrow D=\dfrac{11}{45}\)